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Let $$\alpha(x) = \Biggl\{ \begin{array}{cc} 0 & x=0, \\\ 2 & 0<x\leq 1 \\\ 1 & 1<x \leq 2 \end{array}$$

Let $f(x)$ be continuous on $[0,2]$. Let $P = \{0,0.5,1,1.5,2\}$. For this partition, find the sum $S(P,t)$ approximating the integral

$$\int\limits_{0}^{2} f(x) d \alpha(x)$$

for every $t$ for which this sum is defined. What is the value of the integral? Justify your answer by referring to the definition of the integral only.

I know the definition but cannot seem to apply to it to cases. This is bad, I know, but can you go through this example with me. Once I understand how to apply it I'm sure I'll be able to figure out the rest of the questions! :) Oh... I did draw $a(x)$. It's the whole finding $L(P)$ and $U(P)$ with sums that I find difficult.

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You could start with writing out and thinking about the definitions of $L(P)$ and $U(P)$. "I know the definition" is too vague a thing to say. If you've understood the construction of the Riemann integral, then I cannot possibly imagine why this weighted version should cause any trouble whatsoever, especially not in such a simple case. Having a peek at the corresponding Wikipedia page might also help. –  t.b. May 21 '11 at 19:11
@all: if my previous comment should seem overly harsh to anyone among you and you should happen to object to that, please take into account the flood of questions that were asked by the same user yesterday and today. –  t.b. May 21 '11 at 19:20
If you can't apply the definitions, then you don't know them in any meaningful sense. (Not to be harsh, but this usage of the word "know" is a pet peeve of mine. If you can't act on the definitions, then they're just words.) –  Qiaochu Yuan May 21 '11 at 19:59
@Theo: I noticed the same pattern; clearly, I've been fairly vocal about "inappropriate harshness" when addressing users; but rest assured, given this pattern, I have no objections here. –  amWhy May 21 '11 at 20:55

1 Answer 1

up vote 2 down vote accepted

A little hint: You need to apply the definition :-) If you say you cannot to that, I have to agree with Theo and Qiaochu that this is a strong indication that you either don't know, or don't understand, the definition.

In that case, you should try to understand the most simple example that was mentioned in your class. For every definition in every math class, you should come up with the most simple example possible, and some nontrivial example.

The first step in applying the definition is a choice of a partition of the integration interval. Choose some numbers $x_1, x_2, x_3 \in [0, 2]$ and write the sum that approximates the integral. Then choose a refinement of this partition and write down the sum again. Maybe you'll notice that it is possible to choose a sequence of partitions such that you can evaluate the sums and determine the limit that is the integral...

Addendum: Ok, here is another hint: For $\alpha$ differentiable, we know that $$ \int f(x) d \alpha(x) = \int f(x) \alpha'(x) dx $$ So, if $\alpha$ is a jump function, that is constant with the notable exeption of a jump, it is possible to simplify the integral. For example, define $$ \alpha(x) := 0 \; \text{for} \; x \lt 0 $$ and $$ \alpha(x) := 1 \; \text{for} \; x \ge 0 $$
Then we can rewrite the integral $$ \int_{-a}^{a} f(x) d\alpha(x) = \int_{-a}^{-\epsilon} f(x) \alpha'(x) dx + \int_{-\epsilon}^{\epsilon} f(x) d\alpha(x) + \int_{\epsilon}^{a} f(x) \alpha'(x) dx $$ for an arbitrarily small $\epsilon \gt 0, \epsilon \lt a$. Since the differential of $\alpha$ vanishes for the first and the last term, we get $$ \int_{-a}^{a} f(x) d\alpha(x) = \int_{-\epsilon}^{\epsilon} f(x) d\alpha(x) $$ So, it is actually enough to consider what happens in an $\epsilon$ neighborhood of 0, in order to calculate the integral.

Now, any summand in U(P) and in L(P) will be zero except the one containing the difference of $\alpha(\epsilon) - \alpha(-\epsilon)$, so we have $$ U(P) = (\alpha(\epsilon) - \alpha(-\epsilon)) \sup_{ x \in [-\epsilon, \epsilon]} f(x) = \sup_{ x \in [-\epsilon, \epsilon]} f(x) $$ and $$ L(P) = (\alpha(\epsilon) - \alpha(-\epsilon) ) \inf_{ x \in [-\epsilon, \epsilon]} f(x) = \inf_{ x \in [-\epsilon, \epsilon]} f(x) $$ Since $f$ is continuous, we get in the limit $\epsilon \to 0$: $$ U(P) = L(P) = f(0) $$

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Thanks for the comments. In terms of ‘knowing’ the definition – i.e. – I know the words and can apply it to specific cases to work out the integral. For example, for the problem above, I got 2f(0)-f(1) as the value of the integral. I drew alpha(x) and then calculated the jumps. These types of problems are not a problem. I have tried looking in textbooks and stuff to calculate L(P) and U(P) but I just do not understand at all. –  user4645 May 22 '11 at 12:54
I have looked in lecture notes but they just use the definition and magically seem to get everything that follows. It’s not clear and I have asked friends, lecturers, and everyone else I can think of to explain it – I still don’t get how they do it. Maybe I’m just stupid but if you could go through this example with me then I’m more likely to get it then if someone goes through the theory…. And the reason why there’s so many questions in one go is because I’m preparing for exams so have limited time online. –  user4645 May 22 '11 at 12:54
Here’s my attempt anyway: L(P) = sum from 0 to 2 of (alpha(xi)-alpha(xi-1)) inf(f(x)), so it’s: alpha(0.5)-alpha(0)*f(0) + alpha(1)-alpha(0.5)*f(0.5) + alpha(1.5)-alpha(1)*f(1) + alpha(2)-alpha(1.5)*f(1.5). U(P) = sum from 0 to 2 of (alpha(xi)-alpha(xi-1)) sup(f(x)), so it’s: alpha(0.5)-alpha(0)*f(0.5) + alpha(1)-alpha(0.5)*f(1) + alpha(1.5)-alpha(1)*f(1.5) + alpha(2)-alpha(1.5)*f(2). And L(P)<=I<=U(P). This is what I really don’t get. –  user4645 May 22 '11 at 12:55
Sorry, I don't understand what "this" is in your statement "This is what I really don't get". –  Tim van Beek May 22 '11 at 18:26
BTW, when learning mathematics, it is very important to actively work on solving problems all the time, a binge session just before the exams won't work. –  Tim van Beek May 22 '11 at 18:39

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